Problem 18 - Algebra A Combined Function

Question

Solve each system of equations by the substitution method. $$ \begin{array}{l} 2 y=x+2 \\ 6 x-12 y=0 \end{array} $$

Step-by-Step Solution

Answer

The system has no solution as it results in a contradiction (-12 = 0).

Step 1: Solve for 'y' in the first equation

Given the equation \(2y = x + 2\), we solve for \(y\) by dividing both sides by 2: \[y = \frac{x + 2}{2}\].

Step 2: Substitute 'y' in the second equation

Replace \(y\) in the second equation \(6x - 12y = 0\) with the expression found in Step 1: \(y = \frac{x + 2}{2}\). Then the equation becomes: \[6x - 12\left(\frac{x + 2}{2}\right) = 0\].

Step 3: Simplify the substituted equation

Distribute the 12 inside the parentheses: \[6x - 12\left(\frac{x}{2} + 1\right) = 0\] becomes \[6x - 6x - 12 = 0\].

Step 4: Solve the equation for 'x'

Simplify the equation: \(-12 = 0\). Since this is a contradiction, the system of equations has no solution.

Key Concepts

System of Equations

In algebra, a system of equations is a collection of two or more equations that share the same set of variables. The goal is to find values for these variables that satisfy all the groups of equations simultaneously. When dealing with linear systems, each equation represents a line in a two-dimensional space. Thus, finding a solution to a system of equations means locating the point at which the lines intersect.

The given problem consists of two equations: \(2y = x + 2\) and \(6x - 12y = 0\). These equations are linear, each representing a straight line on the Cartesian plane. A "solution" in this context would be the coordinates \((x, y)\) of the point where both lines would meet. However, when we explored the solution, it was found that no such intersection occurs, meaning the lines are parallel and therefore, there is no common solution.

This leads us to conclude that the system of equations has no solution, which is an important consideration when analyzing systems. Not all systems will have solutions that satisfy all equations; some might have infinite solutions, while others, like this one, have none at all.

Solving Linear Equations

To solve a system of linear equations using the substitution method, we focus on one equation to express one variable in terms of another. This technique helps to simplify the problem, making it easier to substitute these expressions into the other equations.

In this exercise, we start with the first equation \(2y = x + 2\) to solve for \(y\). By isolating \(y\), we can express it as \(y = \frac{x + 2}{2}\). This provides us a way to substitute \(y\) in terms of \(x\) into the second equation, \(6x - 12y = 0\).

The substitution reduces the number of variables in the second equation from two (\(x\) and \(y\)) to one (only \(x\)). This simplification is key in solving linear equations, as it transforms the system into a form where we can solve for \(x\) directly. Unfortunately, when we simplify in this specific exercise, we reach a contradiction, proving that no solution exists. It's important to recognize these inconsistencies, as they indicate relations between lines such as parallelism.

Algebra Concepts

Key algebra concepts are at play when working with systems of equations. Understanding substitution and simplification are central to making sense of this kind of problem.

• Substitution: This involves replacing one variable in an equation with an equivalent expression from another equation. It helps in reducing the complexity of the system.
• Simplification: After substitution, the next step is to simplify the resulting equation. This often involves distributing coefficients and combining like terms. Simplicity in form often makes solving for variables more straightforward.
• Contradiction and Consistency: A solution might not always exist. In cases like the one given, simplifying leads to a contradiction, pointing us towards the conclusion that the lines do not intersect.

Recognizing these concepts help students not only solve algebraic problems but also understand the nature of the systems of equations they are working with. Each step, from isolating a variable to substituting back into another equation, contributes to an overall understanding of algebraic principles and problem-solving strategies.